Question #272317

Q5. A 5 kg of ice cube at -10°C is mixed with 0.1 kg of water at 80 °C. There is no heat





loss to the surrounding. The specific heat capacity of ice is 2.22 kJK-1kg ? The specific





heat capacity of water is 4.187 kJK-kg“The specific latent heat of fusion of ice is





333 kJ kg-1





(a) What is the final physical state of the mixture? (Gas, Liquid or Solid?)





Justify your answer using suitable calculations





(b) What is the final temperature of the mixture?

Expert's answer

a)

Qitot=cimi(0ti)=111 kJ,Q_{i_{tot}}=c_im_i(0-t_i)=111~kJ,

Qwtot=cwmw(tw0)=33.5 kJ,Q_{w_{tot}}=c_wm_w(t_w-0)=33.5~kJ,

Qitot>Qwtot,Q_{i_{tot}}>Q_{w_{tot}}, it will be an ice,

b)

Qi=Qwtot+Qwλ+Qwi,Q_i=Q_{w_{tot}}+Q_{w_{\lambda}}+Q_{w_i},

cimi(tti)=cwmw(tw0)+λmw+cimw(0t),c_im_i(t-t_i)=c_wm_w(t_w-0)+\lambda m_w+c_im_w(0-t),

t=mw(λ+cwtw)+cimitici(mi+mw)=3.9°C.t=\frac{m_w(\lambda+c_wt_w)+c_im_it_i}{c_i(m_i+m_w)}=-3.9°C.


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